code igniter ::发送url作为param？

Question

我正在创建一个ajax函数来编辑一些设置.

这些必须通过网址处理

(即)

该网址为http://example.com/setting1/setting2/setting3

这很好但是.其中一个设置是网址.

我怎么能以这种方式传递这个？

Answer 1

这可能会奏效

$url = 'http://stackoverflow.com/questions/4245416/code-igniter-send-url-as-param';

$encoded = base64_encode($url);
//aHR0cDovL3N0YWNrb3ZlcmZsb3cuY29tL3F1ZXN0aW9ucy80MjQ1NDE2L2NvZGUtaWduaXRlci1zZW5kLXVybC1hcy1wYXJhbQ==

// Make your URL
$my_url = "site.com/controller/method/params/params/" . $encoded;

// redirect or whatever

$encoded_url = $this->uri->segment(4); // (or wherever the URL is)

$url = base64_decode($encoded_url);
// http://stackoverflow.com/questions/4245416/code-igniter-send-url-as-param